Question

If $y^2=a{x}^2+bx+c$, where $a,b$ and $c$ are constant, then $y^3\frac{d^{2}y}{d{x}^2}$ is

• A. a constant
• B. a function of $x$
• C. a function of $y$
• D. a function of $x$ and $y$ both

Answer 1

The correct option is B

a function of $x$

Explanation for the correct answer:

Differential equation:

$y^2=a{x}^2+bx+c$

Differentiating the given equation with respect to $x$ we get

$2y\times \frac{dy}{dx}=2ax+b$ $...\left(i\right)$

$\Rightarrow$$\frac{dy}{dx}=\frac{2ax+b}{2y}$ $...\left(ii\right)$

Differentiating $\left(i\right)$ with respect to $x$ we get

$2{\left(\frac{dy}{dx}\right)}^2+2y\frac{d^{2}y}{d{x}^2}=2a$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \frac{d}{\mathrm{dx}}\left(u.v\right)=u\frac{\mathrm{dv}}{\mathrm{dx}}+v\frac{\mathrm{du}}{\mathrm{dx}}\right]$

Dividing by $2$ throughout the equation we get

$\Rightarrow$ ${\left(\frac{dy}{dx}\right)}^2+y\frac{d^{2}y}{d{x}^2}=a$

$\Rightarrow$$y\frac{d^{2}y}{d{x}^2}=a-{\left(\frac{dy}{dx}\right)}^2$

$\Rightarrow$$y\frac{d^{2}y}{d{x}^2}=a-{\left(\frac{2ax+b}{y}\right)}^2$ …[From(ii)]

Multiplying by $y^2$ throughout the equation we get

$\Rightarrow$$y^3\frac{d^{2}y}{d{x}^2}=a{y}^2-{\left(2ax+b\right)}^2$

Substituting $y^2=a{x}^2+bx+c$ we get

$\Rightarrow$$y^3\frac{d^{2}y}{d{x}^2}=a\left(a{x}^2+bx+c\right)-{\left(2ax+b\right)}^2$

Hence, $y^3\frac{d^{2}y}{d{x}^2}$ is a function of $x$ only as there are no terms containing $y$ in the RHS,

Hence, option (B) is the correct answer.